What is the change in Gibbs' free energy for #"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"#?

The #∆S^@# for the reaction,

#"C"(s) + "O"_2(g) -> "CO"_2(g) + "393.51 kJ/mol"#,

at #"308.62 K"# is #"0.004 kJ/mol"cdot"K"#.

Find the value of #∆G^@# at #"308.62 K"#.

What would be the answer in #"kJ/mol"#? Thanks!

2 Answers

Follow the steps below

Explanation:

  1. Look at the reaction and determine the #DeltaH^@#
    Since the reaction is exothermic: #DeltaH^@= -393.51 ("kJ")/("mol")#

  2. Given #DeltaS^@ = 0.004 ("kJ")/("mol"*"K")#

  3. Given #T= "308.62 K"#

  4. Use the formula: #DeltaG^@=DeltaH^@-TDeltaS^@#

  5. Plug in givens: #DeltaG^@=-393.51 ("kJ")/("mol")-(308.62 cancel("K"))*(0.004 ("kJ")/("mol"*cancel("K")))#

#= -394.74 ("kJ")/("mol") #

Mar 18, 2018

#-394.74# #" kJ"/"mol"#

Explanation:

To solve this question, use the standard-state free energy of reaction equation:

#Delta"G"_0=DeltaH_0 - TDeltaS_0#

Given that you were given all the values for enthalpy, temperature, and entropy:

#Delta"G"_0=(-393.51 " kJ"/"mol") - (308.62" K")(0.004 "kJ"/("mol" *"K"))#

#Delta"G"_0= -394.74# #" kJ"/"mol"#

Since the value Delta #"G"_0# is negative, this reaction is thermodynamically favorable.